How do I get the mass of water for my thermal energy transfer equation when I wasn't given it?

Question

A 17.27 gram sample of aluminum initially at 92 oC is added to a container containing water. The final temperature of the metal is 25.1 oC. What is the total amount of energy in Joules added to the water? What was the energy lost by the metal? How do I calculate the mass of water? overall I'm just confused

Answer 1

Well, first of all you need to know the specific heat of the aluminum. The heat lost by the aluminum will be equal to the heat gained by the water. We look up the specific heat of Al and find it to be 0.900 J/g^o

Now, heart lost by Al = q = mCΔT

q = (17.27 g)(0.900 J/g^o)(92-25.1)

q = 1040 J. THIS IS THE ENERGY LOST BY THE METAL

The mass of water is determined as follows since heat lost by metal must equal heat gained by water

Heat gained by water = 1040 J

specific heat of water = 4.184 J/g^o

mass water = ?

q = MxΔT

1040 J = (m)(4.184)(uh oh)

since we don’t know the initial temperature of the water we can’t know the change in temperature so we can’t answer this question.